Belt drives systems may be assembled to (Figure 9.1): Opened belt drive (non-reversing) Reversing crossed or twist belt drive Angled belt drive

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1 heory of Machines / Belt Drives 9. Belt Drives 9. Geometry Specification 9.. Construction: Belt drives systems may be consist of (Figure 9.): Driving pulley Driven pulleys (multiple driven sources) Belts Idler and levers (used for adjustment) ω Smaller pulley v belt v Ω Opened belt drive Crossed belt drive Larger pulley Belt drives systems may be assembled to (Figure 9.): Opened belt drive (non-reversing) Reversing crossed or twist belt drive Angled belt drive Reversing open belt drive Angled belt drive Driver Driven Driven Multi output supplies here are many types of belts used these days, these are: Flat belt: used for moderate amount of power is to be transmitted. V-belt: greater amount of power is to be transmitted, and when the two pulleys are very near to each other. Circular Belt or Rope: greater amount of power is to be transmitted, and when the two pulleys are more than 5 m apart. iming belt: It has teeth that fit into groove cut on the periphery of the pulleys. A timing belt dose not stretches or slips and consequently transmits power at a constant angular velocity ratio, Figure 9.. belt Reversing open-belt belt drive drive Figure 9. Figure

2 heory of Machines / Belt Drives 9.. Angle of lap (contact), q: he belts are partly wound round pulleys. he smaller wound angle is called the Angle of lap or angle of contact. a. For opened belt drive: R, r radii of large and smaller pulleys respectively θ angle of lap S distance between pulley's centers Consider the triangle O O P (Figure 9.) O P R - r O O S cos (θ/) (R - r)/s θ cos - (R r)/s θ π/90* cos - (R r)/s in degree in radian If R r θ π rad 80 v P υ r υ/ R O O Ω b. For closed belt drive: Figure 9. Consider the triangle O O P (Figure 9.4) O P R + r O O S cos (ϕ/) (R + r)/s ϕ cos - (R + r)/s θ 60 - cos - (R + r)/s θ π/90* (80 - cos - (R + r)/s) in degree in radian - -

3 heory of Machines / Belt Drives P r θ ω/ R O v O Figure Length of the belt (L): a. For an opened belt drive: L rθ + R*(π - θ) + *(O P); O P S*sin θ/ L rθ + R*(π - θ) + *S*sin θ/ b. For a closed belt drive: L (R + r)θ + *(O P); O P S*sin ϕ/ S* sin (80 - θ/) L (R + r)θ + *S*sin (80 - θ/) 9..4 Speed Ratio of a Simple Belt Drive (e): a. If the larger pulley is the driver: e ω/ω R/r for no slipping and the thickness is not considered b. If smaller pulley is the driver: e Ω/ω r/r [ for no slipping and the thickness is not considered where Ω, and ω is the speed of the larger and smaller pulley in radian per second respectively Speed Ratio of a Compound Belt Drive (e): a. If the larger pulley is the driver and if there are four pulleys in the system: e ω/ω R *R /r *r 4 for no slipping and the thickness is not considered b. If smaller puller is the driver and if there are four pulleys in the system: e Ω/ω r *r /R *R 4 [ for no slipping and the thickness is not considered respectively Slip of belt Sometimes, the frictional grip becomes insufficient. his may cause some forward motion of the driver without carrying the belt with it. his may also cause some - -

4 heory of Machines / Belt Drives forward motion of the belt without carrying the driven pulley with it. his is called slip of the belt and is generally expressed as a percentage. he result of belt slipping is to reduce the velocity ratio of the system. Let s % Slip between the driver and the belt, s % Slip between the belt and the follower. v Velocity of the belt, passing over the driver per minute. d d ( s )( If thickness of the belt (t) is considered, then d d + t ( s + t )( s ) s ) Example : An engine, running at 50 rpm, drives a line shaft on the by means of a belt. he engine pulley is 75 cm diameter and the pulley on the line shaft being 45 cm. A 90 cm diameter pulley on the line shaft drives a 5 cm diameter pulley keyed to a dynamo shaft. Find the speed of the dynamo shaft, when,. there is no slip,. there is a slip of % at each drive. Given 50 rpm, D 75 cm, D 45 cm, D 90 cm, D 4 5 cm. When there is no slip: 4 4 * 4 D D D * D 50* rpm 4 75* 90 45* 5 0. When there is a slip of % at each drive 4 4 * 4 D D D * D ( s 50* rpm 4 )( s ) 75* 90 * ( 45* 5 00 )( 00 ) Example : Find the length of belt necessary to drive a pulley of 80 cm diameter running parallel at a distance of meters from the driving pulley of diameter 480 cm

5 heory of Machines / Belt Drives Given: R 40 cm, r 40 cm, S 00 cm. If the belt is open θ cos - (R r)/s cos - (40 40)/ rad L rθ + R*(π - θ) + *S*sin θ/ 40 * *(π ) +*00* cm. m. If the belt is crossed θ 60 - cos - (R + r)/s rad L (R + r)θ + *S*sin (80 - θ/) 45.8 cm.45 m 9. Ratio of Belt tensions: 9.. Flat Belt: Consider a flat belt partly wound round a pulley so that the angle of lap q, Figure 9.5, and let and be the tensions in the belt when it is about to slip in the direction shown. If the tensions at the ends of an element subtending an angle dq at the center are and + d, the reaction, and the coefficient of friction between the belt and the pulley is R and µ respectively, then resolving forces radially: dθ dθ ( + d ) + herefore, neglecting the second order of small quantities * dθ R Resolving forces tangentially, R (*) ight Side r Figure 9.5 Slack Side Figure

6 heory of Machines / Belt Drives ( + d ) d µ R µ dθ d ln µ dθ d µ θ µ R θ 0 µ dθ e µθ If the belt is used to transmit power between two pulleys, Figure 9.6, and are the tight and slack side tensions respectively. If the pulleys are unequal diameter, the belt will slip first on the pulley having the smaller angle of lap, i.e. on the smaller pulley. If v is the speed of the belt in m/s and and are in ewton, then Power transmitted ( )*v Watt *(-/e µq )*v Watt Example : Find the power transmitted by a belt running over a pulley of 60 cm diameter at 00 rpm. he coefficient of friction between the belt and the pulley is 0.5, angle of lap 60 and maximum tension in the belt is.5 k. Solution Given d 60 cm 0.6 m, 00 rpm, µ 0.5, q rad,.5 k v π*d*00/60 π m/s / e µq k Power transmitted ( )*v (.5.4) * π kw Example 4: wo pulleys, one 450 mm diameter and the mm diameter are on parallel shafts.95 m apart and are connected by crossed belt. Find the length of the belt required. What power the belt can transmit when the larger pulley rotates at 00 rpm, if the maximum permissible tension in the belt is k, and the coefficient between the belt and pulley is 0.5? other 00

7 heory of Machines / Belt Drives Given R 0.5 m, r 0. m, S.95 m, 00 rpm, k, µ 0.5. v π*r*00/ m/s q 60 - cos - (R + r)/s 60 - * cos - ( )/ rad L (R + r)q + *S*sin (80 - q/) ( )* *.95*sin ( ) m / e µq 000/e Power transmitted ( )*v ( )* W.76 kw 9.. Effect of Centrifugal ension: Consider a belt, of mass m per unit length, wound round a pulley of radius r, Figure 9.7. Let the speed of the belt be v and the centrifugal tension be c. Figure 9.7 If F is the centrifugal force acting on an element of the belt subtending an angle dθ at the center, then resolving forces radially, or dθ v F c i.e mrdθ. cdθ r c mv his is the tension caused by centrifugal force on the belt and is additional to the tension due to the transmission of power. Equation (*) due to this additional tension becomes: v dθ R + F R + mrdθ r So that d µ R µ ( - mv ) dθ d i.e. µdθ - c - 7 -

8 heory of Machines / Belt Drives or d c c c θ 0 µθ e µ dθ - c and - c are the effective driving tensions and and are now the total tensions in the belt. he transmitted power with the effects of centrifugal tension is; ( ) ν µ power ( )* v c θ e From the above equation the power transmitted is a maximum when d dv d dv mv c {( ) ν} c ( v mv ) 0 0 e Maximum power * * v µθ opt where v opt 9.. Initial ension: * m he belt is assembled with an initial tension, o. When power is being transmitted, the tension in the tight side increases from o to and on the slack side decreases from o to. o o or + o eglecting centrifugal tension 9. V Belt or Rope Drive 9.. Advantages and Disadvantages of V-belt Drive over Flat Belt Drive Advantages - 8 -

9 heory of Machines / Belt Drives. he V-belt drive gives compactness due to small distance between centers of pulleys.. he slip between the belt and the pulley groove is negligible.. he operation of the belt and pulley is quiet. 4. he high velocity ratio (maximum 0) may be obtained. 5. he power transmitted by V-belts is more than flat belts for the same coefficient of friction, arc of contact (angle of lap) and allowable tension in the belt. 6. he V-belt may be operated in either direction, with tight side of the β belt at top or bottom. he center line may be horizontal, vertical, or inclined. R V-belt Disadvantages. he V-belt drive cannot be used with large center distance.. he V-belts are not as durable as flat belts.. he construction of pulleys for V-belts is more complicated than pulleys of flat belts. β R/ 9.. Ratio of V-belt ensions: As in the flat belt drive, let tight tension in belt slack tension in belt R total reaction in the plane of the groove ormal reaction between belt and sides of the groove β angle of the groove µ coefficient of friction between the belt and sides of the groove From the force polygon in Figure 9.8, R sin β R cosecβ V-grooved pulley Figure 9.8 We have two friction forces, F µ * f \ F µ * R* cos ecβ f he friction force is therefore increased in the ratio cosecβ:, so that the V-grooved pulley is equivalent to a flat pulley having a coefficient of friction of µ.cosecβ. Hence - 9 -

10 heory of Machines / Belt Drives e µθ cos ec β When the effect of centrifugal tension is considered, the tension ratio became: - - c c µθ cos ecβ e Power ( ) * v ( C )* (-/e µθcosecβ )* v For the condition of maximum transmitted power, Maximum power e * * v µθ cosecβ opt where v opt * m Example 5: A belt drive consists of two V-belts in parallel, on grooved pulleys of the same size. he angle of the groove is 0. he cross-sectional area of each belt is 750 mm and µ 0.. he density of the belt material is. Mg/m and the maximum safe stress in the material is 7 M/m. Calculate the power that can be transmitted between pulleys 00 mm diameter rotating at 500 rpm. Find also the shaft speed in rpm at which the power transmitted would be a maximum. Given β 5, θ 80 π rad, µ 0., 500 rpm v π*r*/60.56 m/s We know that mass of the belt per meter length m area x density 750 x 0-6 x kg/m Centrifugal tension, c mv 0.9 x and maximum tension maximum stress x cross-sectional area of the belt 7 x 0 6 x 750 x ( c )/ ( c ) e µθ.cosecβ Power transmitted *( - ) * v 7.68 kw For maximum power, c / 750 mv 750 v (750/0.9) 0.5 v 44. m/s ω*r - 0 -

11 heory of Machines / Belt Drives ω 44./ rad/s 0* 94/π 807 rev/min Example 6: A rope drive is required to transmit 0 kw from a pulley of m diameter running at 450 rpm. he safe pull in each rope is 800 and the mass of the rope is 0.46 kg/m. he angle of lap and the groove angle are 60 and 45 respectively. If the coefficient of friction between the rope and the pulley is 0., find the number of ropes required. v π * d* /60.56 m/s c m* v 0.46 * ( c ) / ( c ) e µθ. cosecβ We know that power transmitted per rope; ( )* v.40 kw o. Of ropes otal power transmitted/ Power transmitted per rope 0/ or Example 7: An open belt drive connects two pulleys. and 0.5 m diameter, on parallel shafts.6 m apart. he belt has a mass of 0.9 kg/m length, and the maximum tension in it is not to exceed k. he. m pulley, which is the driver, runs at 00 rev/min. Due to belt slip on one of the pulleys, the velocity of the driven shaft is only 450 rev/min. Calculate the torque on each of the two shafts, the power transmitted, and the power lost in friction. µ 0.. What is the efficiency of the drive? From Figure 9.0, cos (θ/) ( )/ θ/ 84 5'.47 rad Angle of lap on smaller pulley, θ.994 rad he belt speed is that corresponding peripheral speed of the large (driving) pulley. v 00 * π * 0.6 / 0.57 m/s Figure to the

12 heory of Machines / Belt Drives C m * v 0.9 * c µθ e c 90 orque on driver ( ) * R (000 90) * m orque on follower ( ) * r (000 90) * m Power ( ) * v (000 90) * W.7 kw If there were no slip, speed of follower would be: ω speed ratio * Ω. * 00 / rev/min Power transmitted to follower.7 *450 / kw Power lost in friction kw Efficiency output power/ input power (.85 /.7) * % Example 8: A small air compressor is belt-driven from a lay shaft in a workshop, the pulley on the compressor being 00 mm diameter, and the angle of lap of the belt is 65. When the belt is moved from the loose to the fast pulley, it slips for 8 s until the compressor attains its constant speed of 00 rev/min. he flywheel of the compressor has a moment of inertia of 4 kg m and the friction requires a constant torque of 4 m. If the coefficient of friction is 0.8 during the accelerating period, find the tensions in both reaches of the belt, and also the distance that the belt slips and the energy lost in that time due to belt slip. θ 65 π*65/ rad ω π*00/60.46 rad/s While slip is taking place, ratio of belt tensions, / e µθ e 0.8* () Angular acceleration of compressor α dω/dt.46/8.97 rad/s et torque on compressor ( )* R 4 I * α (4 * ) / () herefore, from () and (): 7 and 06 Belt velocity v ω * R.46 * m/s Distance moved in 8 sec. 8 * m Distance moved by a point on the circumference 0.5 * m Slip of belt relative to pulley m Energy lost due to slip ( )* distance slipped.4 * J - -

13 heory of Machines / Belt Drives Example 9: wo parallel horizontal shafts, whose center lines are 4.8 m apart, one being vertically above the other, are connected by an open belt drive. he pulley on the upper shaft is.05 m diameter that on the lower shaft.5 m diameter. he belt is 50 mm wide and the initial tension in it when stationary and when no torque is being transmitted is k. he belt has a mass of.5 kg/m length; the gravitational force on it may be neglected but centrifugal force must be taken into account. he material of the belt may be assumed to obey Hook's Law, and the free lengths of the belt between pulleys may be assumed to be straight. he coefficient of friction between the belt and either pulley is 0.. Calculate a) he pressure in /m between the belt and the upper pulley when the belt and pulleys are stationary and no torque is being transmitted; b) he tension in the belt and the pressure between the belt and the upper pulley if the upper shaft rotates at 400 rev/min and there is no resisting torque on the lower shaft, hence no power being transmitted; c) he greatest tension in the belt in the belt if the upper shaft rotates at 400 rev/min and the maximum possible power is being transmitted to the shaft. υ 000 p dυ 000 (b) υ/ (a) Figure 9.0 a) Let the pressure on an element subtending an angle dθ at the center, Figure 9.0b, be p /m. hen, resolve forces radially. p * 0.55 * dθ * 0.5 * 000 * sin dθ/ p 800 /m b) v ω*r (π*400/60)* m/s C m*v.5 * Because part of the tension in this case is due to centrifugal tension, however, and the reaction between the belt and the pulley is reduced. - -

14 heory of Machines / Belt Drives Effective tension p 800 * 75/ /m c) cos (θ/) ( )/ θ 74 8'.05 rad + o 6000 () Also c µθ e e 0.*.05.5 () c herefore, from equations () and (), 970 Example 0: A compressor, requiring 90 kw, is to run at about 50 rpm. he drive is by V-belts from an electric motor running at 750 rpm. he diameter of the pulley on the compressor shaft must not be greater than m while the center distance between the pulleys is limited to.75 m. he belt speed should not exceed 600 m/min. Determine the number of V-belts required transmitting the power if each belt has a cross-sectional area of 75 mm, weighs 0.00 kg / cm, and has allowable tensile stress of.5 M / m. he groove angle of the pulley is 5. he coefficient of friction is 0.5. Calculate also the length required of each belt. Ω / ω r / R r 0.5 * 50/ m v 600 / m/s he weight of the belt per length, m ρ * A 0.00 * 0 6 * 75 * kg / m C m * v 0.75 * Maximum tension in the belt,.5 * 0 6 * A.5 * 0 6 * 75 * C For open belt drive, cos (θ/) ( )/ θ rad For V-belt drive, the ratio of belt tension with centrifugal effect taking into account, ( C ) / ( C ) e µθ.cosecβ e 0.5*.757*cosec (7.5) / he power transmitted by one belt, Power ( ) * v ( ) * W 6.94 kw - 4 -

15 heory of Machines / Belt Drives umber of V-belts 90 / belts Length of belt, * * (π -.757) + *.75 * sin m Example : An electric motor running at 400 rev/min transmits power by V-belts, each of 0 mm cross-sectional area, the total angle of groove being 45. he density of the belt material is.65 Mg/m and the maximum allowable working stress in the belts is M/m. µ 0.. he angle of lap on the motor pulley is 45. Calculate the maximum power, which can be transmitted, and the corresponding diameter of the motor pulley. ω π*400/ rad/s θ radians m ρ * A 650 * 0 * kg/m stress * A * 0 6 * 0 * In the maximum power condition, and or L rθ + R*(π - θ) + *S*sin θ/ C / 640 /. v opt ( /m) m/s e µθcosecβ e 0.*.507*cosec(.5).75 max. power (640.) * (-/.75) * 0. * 887 W 8.87 kw 8.9 kw Maximum power * e * v * 640* ( )* 0.*.75 Example : he following 887 W 8.87 kw data refer to an open belt drive: Diameter of larger pulley 400 mm Diameter of smaller pulley 50 mm Distance between two pulleys m Coefficient of friction 0.4 Maximum tension when the belt is on the point of slipping 00 Find the power transmitted at a speed of 0 m/s. It is desired to increase the power. Which of the following two methods you will select?. Increasing the initial tension in the belt by 0 %.. Increasing the coefficient of friction by 0 %. µθ cos ecβ opt * n - 5 -

16 heory of Machines / Belt Drives cos (θ/) (0. 0.5)/ θ radians / e µθ e 0.4* / o + o 776 power ( ) * v (00 5) * W 8.48 kw. power transmitted when the initial tension increased by 0 %: o * e µθ const o ( / +) * / (e µθ +) * power ( ) * v ( ) * W 9.8 kw. power transmitted when the coefficient of friction increased by 0 %: µ new * e 0.44* o 776 ( ) / * 776 / ( ) 9.7 and * power ( ) * v (. 9.7) * 0 96 W 9.6 kw Since the power transmitted by increasing the initial tension is more, therefore, we shall adopt the first method, i.e. increasing the initial tension. Example : A flat belt is required to transmit 5 kw from a pulley of.5 m effective diameter running at 00 rpm. he angle of contact is stretch over /4 of the circumference and the coefficient of friction between belt and pulley surface is 0.. Determine, taking centrifugal tension into account, width of the belt required. It is given that the belt thickness is 9.5 mm, density of its material is. Mg/m and the related permissible working stress is.5 /mm. θ 60 */ rad m ρ * t * b 00 * 9.5 * 0 - * b 0.45b kg/m v π*r*/60.56 m/s C m* v m 580.5b σ max * t * b.5 *0 6 * 9.5*0 - * b 750b Power transmitted: Power ( C )(-/e µθ )* v C Power /[(-/e µθ )* v] b b b b m 4 mm - 6 -

17 heory of Machines / Belt Drives Problems (Belt Drives) Q: wo shafts whose centers are 00 cm apart are connected by a V-belt drive. he driving pulley is supplied with 9.5 kw and has an effective diameter of 0. m. It runs at 000 rpm while the driven pulley runs at 75 rpm. he angle of groove on the pulleys is 40. Permissible tension in 4 cm cross-sectional area belt is. M/m. he material of the belt weighs. g/cm. he coefficient of friction between belt and pulley rim is 0.8. Estimate the number of belts required. (Ans. 0) Q: A leather belt, 5 mm wide and 6 mm thick, transmits power from a pulley 750 mm diameter which runs at 500 rev/min. he angle of lap is 50 and µ 0.. If the mass of m of leather is Mg and the stress in the belt is not to exceed.75 M/m, find the maximum power which can be transmitted. (Ans kw) Q: Power is transmitted between two shafts by a V-belt whose mass is 0.9 kg/m length. he maximum permissible tension in the belt is limited to. k. he angle of lap is 70 and the groove angle 45. If the coefficient of friction between the belt and pulleys is 0.7; find (i) velocity of the belt for maximum power; and (ii) power transmitted at this velocity. (Ans m/s; 0.66 kw) Q4: In a belt drive, the angle of lap of the belt on the small pulley is 50. With a belt speed of 0 m/s and a tension in the tight side of the belt of.5 k, the greatest power which can be transmitted without slip is 0 kw. What increase of power would be obtained for the same belt speed and maximum tension by using an idler pulley to increase the angle of lap to 0? ake into account the centrifugal effect, the mass of the belt being 0.75 kg/m. (Ans..6 kw) Q5: A pulley is driven by a flat belt, the angle of lap being 0. he belt is 00 mm wide by 6 mm thick and has a mass of Mg/m. If µ 0. and the maximum stress in the belt is not to exceed.5 M/m, find the greatest power which the belt can transmit and the corresponding speed of the belt. (Ans kw,.6 m/s) Q6: Power is transmitted between two shafts, 4.5 m apart, by a crossed wire rope passing round two pulleys, of m and m diameter respectively, the groove angle being 40. If the rope has a mass of 4 kg/m, and the maximum working tension is 0 k, determine the maximum power that the rope can transmit, and the corresponding speed of the smaller pulley. µ 0. (Ans kw, 89.9 rev/min) Q7: Power is transmitted from an electric motor to a machine tool by an open belt drive. he effective diameter of the pulley on the motor shaft is 50 mm while that on the machine tool is 00 mm with a center distance of 600 mm. If the motor speed is 440 rev/min and the maximum permissible belt tension is 900, then the maximum power transmissible is 6 kw. It is necessary that the power transmissible be increased to 6.75 kw, using the same pulleys, center distance and motor speed. he belt is treated with a special preparation - 7 -

18 heory of Machines / Belt Drives that increases its coefficient of friction by 0 percent of its existing value, and in addition a jockey pulley may be fitted. Determine, a) the existing coefficient of friction b) the new angle of lap (Ans. 0.9, 95 ) Q8: A belt drive consists of a V-belt working on a grooved pulley, with an angle lap of 60. he cross-sectional area of the belt is 650 mm, the groove angle is 0 and µ 0.. he density of the belt material is Mg/m and its maximum safe stress is 8 M/m of cross-section. Calculate the power that can be transmitted at a belt speed of 5 m/s. (Ans. 79 kw) Q9: he following particulars apply to one pulley of a rope drive between two parallel shafts: Effective diameter of pulley.5 m otal angle of groove 45 Minimum angle of lap 80 Mass of rope per m run 0.45 kg Maximum permitted load per rope 650 Coefficient of friction 0.5 a) Find the power transmitted per rope at a pulley speed of 00 rev/min, if centrifugal tension may be neglected. b) Find the pulley speed when centrifugal tension accounts for half the permitted load in the rope, and the power, which can be transmitted at that speed. (Ans. 8.9 kw, 4 rev/min, 7.6 kw) Q0: A /4 reduction drive between two parallel shafts m centers is provided by means of five parallel V-belts running on suitable pulleys mounted on the shafts. he effective diameter of the driving pulley is 50 mm and the driving shaft rotates at 740 rev/min. he included angle of each pulley groove is 40, each V-belt has a mass of 0.45 kg/m and the coefficient of friction between belt and each groove is 0.8. Determine what power can be transmitted by drive, if the tension in each belt is not to exceed 800. (Ans: 4.9 kw) Q: A small generator is driven by means of a V-belt which has a total angle of 60 between the faces of the V. he angle of lap on pulley is 0 and the mean radius of the belt as it passes round the pulley is 50 mm. If µ 0. and the mass of the belt is 0.45 kg/m, find the tension in each side of the belt when 750 Watt is being transmitted at a pulley speed of 800 rev/min. (Ans: 80 ; 00.5 ) Q: A shaft running at 00 rpm is to drive another shaft at 40 rpm and transmits kw. he distance between the shafts is 50 cm and the smaller pulley is of 60 cm diameter. he flat belt employed is 4 mm wide, mm thick, and the coefficient of - 8 -

19 heory of Machines / Belt Drives friction between the belt and pulley is 0.5. Calculate the stress in the belt if it is (a) an open belt drive; and (b) a cross belt drive